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Thread: inequality problem

  1. #1
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    inequality problem

    For some dang reason I can't figure out this inequality problem..it's just not working! The solution is $\displaystyle c < -2 $ but I keep getting $\displaystyle c > -2$

    Problem: $\displaystyle |\frac{4+c} {c}|<1$
    (absolute value of the fraction)
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  2. #2
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    Quote Originally Posted by sarahh View Post
    For some dang reason I can't figure out this inequality problem..it's just not working! The solution is $\displaystyle c < -2 $ but I keep getting $\displaystyle c > -2$

    Problem: $\displaystyle |\frac{4+c} {c}|<1$
    (absolute value of the fraction)

    $\displaystyle |\frac{4+c} {c}|<1\Longleftrightarrow -1<\frac{4+c}{c}< 1$ , so the next inequalities have both to be true:

    1) $\displaystyle -1< \frac{4+c}{c}\Longrightarrow 2\,\frac{2+c}{c}> 0\Longrightarrow c< -2\,\,\,or\,\,\,c> 0$

    2) $\displaystyle \frac{4+c}{c}< 1\Longrightarrow \frac{4}{c}< 0\Longrightarrow c< 0$

    From (1)(2) we get $\displaystyle c< -2$

    Tonio
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  3. #3
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    Hi Tonio--I still don't understand how you got line (1) and wouldn't line (2) produce $\displaystyle 4 < 0 $ and not $\displaystyle c < 0 $ the former being nonsensical since the variable is gone when multiplied by 0.

    Here's why I come out with $\displaystyle c > -2 $ :

    $\displaystyle |\frac{4+c} {c}|<1$
    $\displaystyle \frac{4}{c} + 1>-1$
    $\displaystyle \frac{4}{c}>-2$
    $\displaystyle (c)\frac{4}{c}>-2c => 4>-2c$
    Flip signs since dividing by a negative! $\displaystyle -2<c => c>-2$
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  4. #4
    Super Member 11rdc11's Avatar
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    1)
    $\displaystyle -1 < \frac{4+c}{c} $

    $\displaystyle 0 < \frac{4+c}{c} +1$

    $\displaystyle 0< \frac{4+2c}{c}$


    now set your denominator and numerator equal to 0

    and you get c equals -2 and 0

    Now test a value to the left of -2 on the number line, test a value between -2 and 0, and then test a value to the right of 0 into the inequality

    $\displaystyle 0< \frac{4+2(-3)}{-3}= \frac{2}{3}$

    which makes the inequality true

    $\displaystyle 0< \frac{4+2(-1)}{-1} = -2$

    which makes the inequality false

    $\displaystyle 0< \frac{4+2(1)}{1)} = 6$

    Which makes the inequality true

    so $\displaystyle c < -2$ or $\displaystyle c>0$

    2)

    $\displaystyle 1> \frac{4+c}{c}$

    $\displaystyle 0> \frac{4+c}{c} -1$

    $\displaystyle 0> \frac{4+c}{c} -\frac{c}{c}$

    $\displaystyle 0> \frac{4}{c}$

    set your numerator and denominator equal to 0 again and you get c = 0

    Now test a value to the left of o and to the right of zero

    $\displaystyle 0> \frac{4}{-1}= -4$

    which is true

    $\displaystyle 0> \frac{4}{1}= 4$

    which is false.

    so

    c<0

    Now putting all this you notice the inequality of

    $\displaystyle c<0$ and $\displaystyle c<-2$ overlap so

    $\displaystyle c<-2$

    is your answer.
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  5. #5
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by sarahh View Post
    Hi Tonio--I still don't understand how you got line (1) and wouldn't line (2) produce $\displaystyle 4 < 0 $ and not $\displaystyle c < 0 $ the former being nonsensical since the variable is gone when multiplied by 0.

    Here's why I come out with $\displaystyle c > -2 $ :

    $\displaystyle |\frac{4+c} {c}|<1$
    $\displaystyle \frac{4}{c} + 1>-1$
    $\displaystyle \frac{4}{c}>-2$
    $\displaystyle (c)\frac{4}{c}>-2c => 4>-2c$
    Flip signs since dividing by a negative! $\displaystyle -2<c => c>-2$
    The problem with trying to work it out this way is when you are multiplying by c to both sides of the equation cancel out the c in the denominator you assumed c is positive. We do not know if c is positive or negative.
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  6. #6
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    Wow 11, I never saw a problem like this in any math book I've read lol. Very different approach to finding the solution, but I get it! Thanks!
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